Question

A charged particle with a velocity $2\times {10}^{3}m{s}^{-1}$ passes undeflected through electric field and magnetic fields in mutually perpendicular directions. The magnetic field is $1.5$T. The magnitude of electric field will be.

• A. $1.5\times {10}^{3}N{C}^{-1}$
• B. $2\times {10}^{3}N{C}^{-1}$
• C. $3\times {10}^{3}N{C}^{-1}$
• D. $1.33\times {10}^{3}N{C}^{-1}$

Answer 1

Answer: (C)

Given : $v=2\times {10}^3$ m/s $B=1.5T$

As the particle passes undeflected, thus electric force is balanced by the magnetic force acting on the particle.

$\therefore$ $qE=q(v\times B)$

$⟹$ $E=Bv=1.5\times 2\times {10}^3=3\times {10}^3$ $N{C}^{-1}$